Question

All chords of the curve $3x^2-y^2-2x+4y=0$ which subtend a right angle at the origin pass through

• A. $(1,-2)$
• B. the point of intersection of the lines $y+2x=0$ and $x=1$
• C. the vertex of the parabola $x^2-2x-4y-7=0$
• D. centre of the circle $x^2+y^2+2x-4y-4=0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A the point of intersection of the lines $y+2x=0$ and $x=1$
B $(1,-2)$
C the vertex of the parabola $x^2-2x-4y-7=0$

Let $y=mx+c$ be a chord of the given curve.

Equation of the pair of lines through the origin and the points of intersection of the chords and the curve is

$3x^2-y^2-(2x-4y)\left(\frac{y-mx}{c}\right)=0$

If these are at right angles

$3+\frac{2m}{c}+(-1+\frac{4}{c})=0$

$\Rightarrow m=-(c+2)$

So the equation of the chord is

$y+(c+2)x-c=0\Rightarrow (y+2x)+c(x-1)=0$.

Now, since the chord passes through the point of intersection of the lines

$y+2x=0$ and $x-1=0$

$\Rightarrow$ The chord passes through the point $(1,-2)$.
Equation of the parabola in (c) is ${(x-1)}^2=4(y+2)$

Vertex is also $(1,-2)$.

Note that the center of the circle in (d) is $(-1,2)$. Hence, not a correct option.

Hence, options A, B, and C.